If  $$a\sin A+b\cos A=c\ $$, then  $$a\cos A-b\sin A$$  is equal to:

$$a\sin A+b\cos A=c\ $$

$$=$$>  $$\left(a\sin A+b\cos A\right)^2=c^2\ $$

$$=$$>  $$a^2\sin^2A+b^2\cos^2A+2ab\sin A\cos A=c^2\ $$

$$=$$>  $$a^2\sin^2A+b^2\left(1-\sin^2A\right)+2ab\sin A\cos A=c^2\ $$

$$=$$>  $$\left(a^2-b^2\right)\sin^2A+b^2+2ab\sin A\cos A=c^2\ $$

$$=$$>  $$\left(a^2-b^2\right)\sin^2A+2ab\sin A\cos A=c^2\ -b^2$$ ..........(1)

Let $$a\cos A-b\sin A=x$$

$$=$$>  $$\left(a\cos A-b\sin A\right)^2=x^2$$

$$=$$>  $$a^2\cos^2A+b^2\sin^2A-2ab\cos A\sin A=x^2$$

$$=$$>  $$a^2\left(1-\sin^2A\right)+b^2\sin^2A-2ab\cos A\sin A=x^2$$

$$=$$>  $$a^2+\left(b^2-a^2\right)\sin^2A-2ab\cos A\sin A=x^2$$

$$=$$>  $$a^2-\left[\left(a^2-b^2\right)\sin^2A+2ab\sin A\cos A\right]=x^2$$

$$=$$>  $$a^2-\left[c^2-b^2\right]=x^2$$

$$=$$>  $$a^2-c^2+b^2=x^2$$

$$=$$>  $$x=\sqrt{a^2+b^2-c^2}$$

$$\therefore\ a\cos A-b\sin A=\sqrt{a^2+b^2-c^2}$$

Hence, the correct answer is Option A